AMC 10 · 2010 · #6
Grade 8 geometry-2dA circle is centered at O, AB is a diameter and C is a point on the circle with ∠COB=50∘. What is the degree measure of ∠CAB?
Pick an answer.
AMC 10 2010 problem © Mathematical Association of America (MAA AMC). Reproduced for educational use.
Try it yourself first — the explanation is most useful after you’ve attempted it.
Toolkit + CCSS Solution
Understand
Restated: A circle has center $O$, and $\overline{AB}$ is a diameter, so $A$, $O$, and $B$ lie on one straight line. A third point $C$ sits on the circle, and the central angle $\angle COB$ measures $50^\circ$. We want the measure of $\angle CAB$, the angle at vertex $A$ inside triangle $ABC$.
Givens: $O$ is the center of the circle.; $\overline{AB}$ is a diameter, so $A$, $O$, $B$ are collinear and $OA$, $OB$, $OC$ are all radii.; $C$ is a point on the circle with $\angle COB = 50^\circ$.; Answer choices: (A) $20$, (B) $25$, (C) $45$, (D) $50$, (E) $65$.
Unknowns: The degree measure of $\angle CAB$.
Understand
Restated: A circle has center $O$, and $\overline{AB}$ is a diameter, so $A$, $O$, and $B$ lie on one straight line. A third point $C$ sits on the circle, and the central angle $\angle COB$ measures $50^\circ$. We want the measure of $\angle CAB$, the angle at vertex $A$ inside triangle $ABC$.
Givens: $O$ is the center of the circle.; $\overline{AB}$ is a diameter, so $A$, $O$, $B$ are collinear and $OA$, $OB$, $OC$ are all radii.; $C$ is a point on the circle with $\angle COB = 50^\circ$.; Answer choices: (A) $20$, (B) $25$, (C) $45$, (D) $50$, (E) $65$.
Plan
Primary tool: #4 Introduce a Variable
Secondary: #1 Draw a Diagram, #7 Identify Subproblems
The angle we want, $\angle CAB$, is one of the two base angles of triangle $OAC$. The clean move is Tool #4 (Introduce a Variable): call that base angle $x$. Because $OA$ and $OC$ are both radii, triangle $OAC$ is isosceles, so the OTHER base angle is also $x$ for free — one variable captures two angles. Tool #1 (Draw a Diagram) makes the picture do the heavy lifting: drawing radius $OC$ reveals the isosceles triangle, and seeing $A$, $O$, $B$ on one line reveals the straight-angle relationship. Tool #7 (Identify Subproblems) splits the job into two easy pieces — first pin down $\angle AOC$ from the straight line, then use the triangle's angle sum to solve for $x$.
Execute — Answer: B
8.G.A.5 Step 1 Name the base angle x
- Draw radius $OC$.
- Now look at triangle $OAC$.
- Sides $OA$ and $OC$ are both radii of the same circle, so $OA = OC$ and the triangle is isosceles.
- In an isosceles triangle the two angles opposite the equal sides are equal, so the angle at $A$ equals the angle at $C$.
- Let $x = \angle CAB$; then $\angle OCA = x$ as well.
💡 Two equal radii force the two base angles to match, so a single letter $x$ stands for both of them.
7.G.B.5 Step 2 Use the straight line at O
- Since $\overline{AB}$ is a diameter, $A$, $O$, $B$ lie on one straight line.
- That means $\angle AOC$ and $\angle COB$ are adjacent angles that together fill the straight angle at $O$, so they are supplementary.
- With $\angle COB = 50^\circ$, this gives $\angle AOC = 180^\circ - 50^\circ = 130^\circ$.
💡 A straight line is a $180^\circ$ angle, so the two pieces on either side of $OC$ must add back up to $180^\circ$.
8.G.A.5 Step 3 Solve the angle-sum equation
- The three angles of triangle $OAC$ add to $180^\circ$.
- Two of them are the base angles $x$, and the third is $\angle AOC = 130^\circ$.
- So $x + x + 130^\circ = 180^\circ$, which gives $2x = 50^\circ$ and $x = 25^\circ$.
- Therefore $\angle CAB = 25^\circ$, which is answer $\textbf{(B)}\ 25$.
💡 Once the top angle is known, the leftover $50^\circ$ splits evenly between the two equal base angles.
8.G.A.5 Draw radius $OC$. Now look at triangle $OAC$. Sides $OA$ and $OC$ are both radii 7.G.B.5 Since $\overline{AB}$ is a diameter, $A$, $O$, $B$ lie on one straight line. Tha 8.G.A.5 The three angles of triangle $OAC$ add to $180^\circ$. Two of them are the base Review
Reasonableness: Check against the inscribed-angle idea. $\angle CAB$ is an inscribed angle standing on arc $BC$, and $\angle COB$ is the central angle standing on the same arc. An inscribed angle is always half of the central angle on the same arc, so $\angle CAB = \tfrac{1}{2}(50^\circ) = 25^\circ$ — exactly what we found. It is also sensible that $\angle CAB = 25^\circ$ is smaller than the central angle $50^\circ$. This matches (B).
Alternative: Use the exterior-angle theorem directly. For triangle $OAC$, the angle $\angle COB$ is the exterior angle at vertex $O$ (since $B$ is the extension of side $AO$ past $O$). An exterior angle equals the sum of the two remote interior angles, so $\angle COB = \angle OAC + \angle OCA = x + x = 2x$. Then $2x = 50^\circ$ gives $x = 25^\circ$ in one line, no straight-angle step needed.
CCSS standards used (min grade 8)
8.G.A.5Use informal arguments to establish facts about angle sum and exterior angles (Arguing that the isosceles triangle's base angles are equal and that the three angles of triangle $OAC$ sum to $180^\circ$ to solve $2x + 130^\circ = 180^\circ$.)7.G.B.5Use facts about supplementary, complementary, vertical, and adjacent angles (Recognizing that $\angle AOC$ and $\angle COB$ are supplementary along the diameter, giving $\angle AOC = 130^\circ$.)
⭐ Two radii make an isosceles triangle, so its two bottom angles are equal; find the top angle from the straight line, and the leftover splits in half to give $25^\circ$.
⭐ Two radii make an isosceles triangle, so its two bottom angles are equal; find the top angle from the straight line, and the leftover splits in half to give $25^\circ$.
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